# Assuming that the range of sin^(-1)x  is (-oo, oo), is  x sin^(-1) x  differentiable, for sin^(-1)x in [0, 2pi]?

Oct 7, 2016

#### Explanation:

For multivalued $y = x {\sin}^{-} 1 x$ we can use the equations

$\frac{y}{x} = {\sin}^{-} 1 x$

$\sin \left(\frac{y}{x}\right) = x$

Implicit differentiation gives:

$\cos \left(\frac{y}{x}\right) \left(\frac{y ' x - y}{x} ^ 2\right) = 1$

$x y ' \cos \left(\frac{y}{x}\right) - y \cos \left(\frac{y}{x}\right) = {x}^{2}$

$y ' = \frac{{x}^{2} + y \cos \left(\frac{y}{x}\right)}{x \cos \left(\frac{y}{x}\right)}$

Replacing $y = x {\sin}^{-} 1 x$ and $\frac{y}{x} = {\sin}^{-} 1 x$ gets us

$y ' = \frac{x + {\sin}^{-} 1 x \cos \left({\sin}^{-} 1 x\right)}{\cos} \left({\sin}^{-} 1 x\right)$

I am uncertain about rewriting $\cos \left({\sin}^{-} 1 x\right)$ in this situation.

It looks to me like the derivative will exist provided that $x \ne \pm 1$

For what it's worth, here is Socratic's graph of $\sin \left(\frac{y}{x}\right) = x$ (I have not restricted values of $x$ or $y$ in any way.)
It's hard to see, but the graph is a sequence of loops getting longer and "more perpendicular". (I just invented that technical description.)

graph{sin(y/x)=x [-1.92, 2.406, -0.684, 1.478]}

Oct 8, 2016

See the explanation

#### Explanation:

What I mean by my assumption that ${\sin}^{- 1} x \in \left(- \infty , \infty\right)$ is as

follows.

If $y = f \left(x\right) = \sin x$, then the graphs of the wave

$y = \sin x$, in infinitude, and the the graph of the inverse x =

sin^(-1)y are one and the same.

And then, for the locally bijective f(x), ${f}^{- 1} f \left(x\right) = x$, everywhere.

Accordingly, it is indubitable that ${\sin}^{- 1} \sin \left(k \pi\right) = k \pi = 0$,

only when k = 0...

I am aware that some students, at Middle School Level, might find

this difficult to follow. So, I use in my answer the conventional

${\sin}^{- 1} x \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$. I have chosen to define y piecewise for the

chosen problem.

Here, ${\sin}^{- 1} x \in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right] , \left({\sin}^{- 1} x\right) ' = \frac{1}{\sqrt{1 - {x}^{2}}}$, with $| x | \le 1$

Using the notion of the general value of inverse sine, define

$0 \le y \le \frac{\pi}{2} , y = x {\sin}^{- 1} x$. x increases from 0 to 1

$- \frac{3}{2} \pi \le y \le \frac{\pi}{2} , y = x \left(\pi - {\sin}^{- 1} x\right)$. x decreases from 1 to $- 1$

$- \frac{3}{2} \pi \le y \le 0 , y = x \left(2 \pi + x {\sin}^{- 1} x\right)$. x increases from $- 1$ to 0

The graph comprises two loops with origin as the common point.

The loops are in between the parallels $x = \pm 1$

Output from $a d$ $h o c$ computer program reveals that the first

loop is in ${Q}_{1}$. It rises from O, touches x= 1 at $\left(1 , \frac{\pi}{2}\right)$, rebound

to the turning point (0.8970, 1.820) at a higher level, for y'=0. .

The second loop is in ${Q}_{3}$.. The arm of of this loop is longer

For my piecewise definition of y, y is continuous and differentiable

everywhere, including the points at which two neighboring pieces

meet.

It will be a graph in grandeur, with loops getting elongated, in

infinitude, in between $x = \pm 1$. It will be is a kind of oscillation for y,

with amplitude increasing without limit...

I welcome a graph for the data given below, for just two loops.
(x, y): (0, 0) (.2, .0427) (.4, .6146) (.6, .3861) (.8, .7418) (1, 1.571) (8, 1.771)
(.6, 1.499) (.4, 1.092) (.2, 0.580) (0, 0) (-.2, 0.6686) (-.4, -1.421) (-.6, -2.271) ( -.8, -3.255) (-1, -4.712) (-.8, -4.285 (-.6, -3.384) (-.4, -2.349), (-.2, -1.216) (0, 0)

Oct 8, 2016

I can't get the Socratic graphing utility on one graph, but here are the two loops.

#### Explanation:

In the first graph, $0 \le x \le 1$ and ${\sin}^{-} 1 x$ is multi-valued in $\left[0 , \pi\right]$

In the second graph, $- 1 \le x \le 0$ and ${\sin}^{-} 1 x$ is multi-valued in $\left[\pi , 2 \pi\right]$
graph{(y-xarcsin(x))(sqrt(x-x^2)/sqrt(x-x^2))(y-x(pi-arcsinx))=0 [-3.385, 4.414, -0.96, 2.937]}

graph{(y-x(pi-arcsinx))(y-x(2*pi + arcsinx))sqrt(x^2-x)/sqrt(x^2-x)=0 [-4.273, 6.83, -5.08, 0.47]}

Using the Desmos graphing website, I got this graph using

$y = x {\sin}^{-} 1 x \text{ for } 0 \le x \le 1$ $\textcolor{red}{\text{ RED}}$
$y = x \left(\pi - {\sin}^{-} 1 x\right) \text{ for } 0 \le x \le 1$ $\textcolor{b l u e}{\text{ BLUE}}$
$y = x \left(\pi - {\sin}^{-} 1 x\right) \text{ for } - 1 \le x \le 0$ $\textcolor{g r e e n}{\text{ GREEN}}$
$y = x \left(2 \pi + {\sin}^{-} 1 x\right) \text{ for } - 1 \le x \le 0$ $\textcolor{p u r p \le}{\text{ PURPLE}}$

Unfortunately these will not zoom or scroll here.

I built them at https://www.desmos.com/calculator